You visited us 1 times! Enjoying our articles? Unlock Full Access!

Wave in Negative X-Direction

The equation ...

Question

The equation of a transverse wave is given by $y = 0.05 sin π (2 t - 0.02 x)$ , where x, y are in metre and t is in second. The minimum distance of separation between two particles which are in phase and the wave velocity are respectively __________.

50 m, 50 m s^{- 1}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

100 m, 100 m s^{- 1}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

50 m, 100 m s^{- 1}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

100 m, 50 m s^{- 1}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $100 m, 100 m s^{- 1}$
The equation of given transverse wave is $y = 0.05 sin π (2 t - 0.02 x)$ .......(1)
On comparing (1) with $y = A sin (w t - k x)$
We get, $k = 0.02 π$ and $w = 2 π$
Wave velocity $v = \frac{w}{k} = \frac{2 π}{0.02 π} = 100$ $m / s$
Minimum distance of separation between two particles which are in phase is equal to the wavelength of the wave.
Wavelength $λ = \frac{2 π}{k} = \frac{2 π}{0.02 π} = 100$ $m$

Suggest Corrections

Similar questions

y = 0.05 sin 2 π (\frac{t}{0.002} - \frac{x}{0.1})

x

y

t

y = 0.05 sin (200 t - \frac{x}{2})

x, y

t

100 m s^{- 1}

10 m s^{- 1}

4 π m

ν = 100 m s^{- 1}

50 m s^{- 1}

100 m s^{- 1}

5 m s^{- 2}

50 m s^{- 1}

100 m

20 m s^{- 1}

Join BYJU'S Learning Program